Example 2
If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.
Given: sin B = sin Q
To prove: ∠ B = ∠ Q
Proof:
Let’s take two right angle triangles ABC & PQR
Since, sin B = sin Q
sin B = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝐵)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
sin B = 𝑨𝑪/𝑨𝑩
sin Q = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑄)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
sin Q = 𝑷𝑹/𝑷𝑸
𝐴𝐶/𝑃𝑅=𝐴𝐵/𝑃𝑄
Let 𝑨𝑪/𝑷𝑹=𝑨𝑩/𝑷𝑸= k
So, AC = k PR & AB = k PQ
Now,
Using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
In right Δ ABC,
(AB)2 = AC2 + BC2
BC2 = AB2 – AC2
BC = √(𝐴𝐵2 −𝐴𝐶2)
In right Δ PQR,
(PQ)2 = (PR)2 + (QR)2
QR2 = PQ2 – PR2
QR = √(𝑃𝑄2−𝑃𝑅2)
Now,
𝑩𝑪/𝑸𝑹 = √(𝑨𝑩^𝟐 − 𝑨𝑪^𝟐 )/√(𝑷𝑸^𝟐 − 𝑷𝑹^𝟐 )
From (1)
AB = k(PQ) and AC = k(PR)
From (1) and (2)
𝐴𝐶/𝑃𝑅 = 𝐴𝐵/𝑃𝑄 = 𝐵𝐶/𝑄𝑅 = k
𝑨𝑪/𝑷𝑹 = 𝑨𝑩/𝑷𝑸 = 𝑩𝑪/𝑸𝑹
Since corresponding sides of
Δ ABC & Δ PQR are in the same ratio
Thus, ∆ ABC ~ ∆ PQR
Since corresponding angles of similar triangles are equal
∴ ∠ B = ∠ Q
Hence proved
𝐵𝐶/𝑄𝑅 = √((𝑘𝑃𝑄)^2 − (𝑘𝑃𝑅)^2 )/√(𝑃𝑄^2 − 𝑃𝑅^2 )
𝑩𝑪/𝑸𝑹= √(𝒌^𝟐 𝑷𝑸^𝟐− 𝒌^𝟐 𝑷𝑹^𝟐 )/√(𝑷𝑸^𝟐 −𝑷𝑹^𝟐 )
𝐵𝐶/𝑄𝑅 = (𝑘√(𝑃𝑄^2 −𝑃𝑅^2 ))/√(𝑃𝑄^2 −𝑃𝑅^2 )
𝑩𝑪/𝑸𝑹 = k
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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